Solving the square of inequality X - 2aX + A + 2 ≤ 0

Solving the square of inequality X - 2aX + A + 2 ≤ 0

x²-2ax+a+2=x²-2ax+a²-(a²-a-2)=(x-a)²-(a²-a-2)≤0
So (x-a) &# 178; ≤ A & # 178; - A-2
When a-178; - A-2
x^2-2ax+a+2=2,a
The inequality that the square of x-2ax + A-1 is less than 0
Do you want to multiply the cross first
(x-a)²-1
If the image of the function y = f (x) in the interval [0,4] is a continuous curve, and the equation f (x) = 0, there is only one real root in (0,4), then the value of F (0) f (4) is zero
A greater than 0 b less than 0 C equal to 0 d cannot be determined
If the image of the function y = f (x) on the interval [0,4] is a continuous curve, then
The function y = f (x) exists on x = 0 or x = 4, but it is uncertain whether it is greater than or less than or equal to 0
The equation f (x) = 0 has only one real root in (0,4), which only means that there is only one intersection point between the curve and X axis in (0,4), and there is no correlation with F (0) f (4);
So the value of F (0) f (4) is uncertain
It's all possible to choose D
Y = (x-1) ^ 2, is a case
Y = X-1 is the case of B
Y = x ^ 2-x is the case of C
Choose B, f (0) < 0, f (4) > 0; or F (0) > 0, f (4) < 0
When x > 1, find the range of F (x) = x & # 178; - 3x + 1 / x + 1
(x (x) is (x ^ 2-2-3x + 1) / (x + 1) / (x + 1) / (x + 1) / (x + 1) (x ^ 2-2-3x-2-3x-3 (x + 2-2-3-3x + 1) as (x + 2-2-2-5) / (x + 2-2-3x + 1) / (x + 1) / (x + 1) is (x) < 0, f (x) monotonmonotonmonotone reduction (x (x) is decreasing (x (x ∈ (- 1 + {(- 1 + {(- 1 +) ((((((((1 \\\\\\\\\ (x) when we (x) min = -1-1-1 = -1-1-1-1 + √ 5-4-4-4-4-4-4 5 range [- 5 + 2 √ 5, + ∞)
It is known that the set M is a set of functions f (x) satisfying the following properties: in the domain of definition, the equation f (x + 1) = f (x) + F (1) has a real solution. (1) does the function f (x) = LX belong to the set M? (2) Let f (x) = lgtx2 + 1 ∈ m, find the value range of T
(1) In the definition domain, \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\2-4 (T-2) × 2 (t-1) ≥ 0 − 6T + 4 ≤ 0 {t ∈ [3 − 5, 2) ∪ (2, 3 + 5]. ｜ t ∈ [3 − 5, 3 + 5]. (12 points)
Given f (x) = Log1 / 2 (- X & # 178; + 3x + 4), find the range of F (x)
As shown in the picture,
-x²+3x+4
=-1/4(4x^2-12x+9-25)
=25/4-(2x-3)^2 /4
Then 0
Let f (x) = | LG | X-1 |, X ≠ 1; 0, x = 1, if the equation a [f (x)] 2-F (x) + 1 = 0 of X has 8 different real solutions
Find the value range of a
When k > 0, f (x) = k has four different real roots,
So if a [f (x)] ² - f (x) + 1 = 0 has 8 different real solutions,
The positive solutions of F are different,
So ⊿ = 1-4a > 0, 1 / a > 0, thus 0
The range of F (x) = 6x & # 178; + 3x + 1 where x belongs to - 2 ≤ x ≤ 2?
f(x)=6x²+3x+1
f'(x)=12x+3
f'(x)=0 x=-1/4
f(-1/4)=5/8
f(2)=31
f(-2)=19
be
The maximum value is 31 and the minimum value is 5 / 8
The range of F (x) on [- 2,2] is: [5 / 8,31]
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The range of F (x) = 6x & # 178; + 3x + 1 where x belongs to - 2 ≤ x ≤ 2 is [5 / 8,31]
The solution is as follows: using derivative
f(x)=6x²+3x+1
f'(x)=12x+3
f'(x)=0 x=-1/4
f(-1/4)=5/8
f(2)=31
f(-2)=19
be
The maximum value is 31 and the minimum value is 5 / 8
The range of F (x) on [- 2,2] is: [5 / 8,31]
When the opening is upward, the function value far from the axis of symmetry is larger, and the function value at the axis of symmetry is smallest
Let g (x) be the derivative of F (x) = 6x & # 178; + 3x + 1,
Let g (x) = 0, and the solution is x = - 1 / 4.
So f (x) is the minimum at x = - 1 / 4, f (- 1 / 4) = 5 / 8;
f(2)=31；f(-2)=19。
Therefore, the range of F (x) = 6x & # 178; + 3x + 1 where x belongs to - 2 ≤ x ≤ 2 is (5 / 8, 31).
It is known that the quadratic function f (x) = ax ^ 2 + BX + C (a, B, C ∈ R) satisfies f (1) = 1, f (- 1) = 0 and has f (x) for any real number X
All have f (x) ≥ X
(1) It is proved that a > 0 C > 0 (2) Let G (x) = f (x) - MX (m ∈ R) find the value of m such that G (x) is monotone on [0,1]
If that's OK, the answer should be more standard
Sub question (1) f (1) = a + B + C = 1F (- 1) = A-B + C = 0, subtracting the two formulas to get b = 1 / 2, so there is a + C = 1 / 2F (x) = ax ^ 2 + (1 / 2) x + (1 / 2 - a) any real number x has f (x) ≥ x, that is, ax ^ 2 - (1 / 2) x + (1 / 2 - a) ≥ 0, the maximum point of intersection with X axis has a > 0, Δ = (1 / 4) - 4a (1 / 2 - a) ≤ 0, that is, a >
（1）f(1)=a+b+c=1,
f(-1)=a-b+c=0.
By subtracting, we get 2B = 1, B = 1 / 2
∴a+c=1/2.(1)
For any real number x, f (x) > = x,
ax^2-x/2+c>=0,
a> 0, and 1 / 4-4ac0, C > 0
(2) By (1), (2), 1 / 4-4a (1 / 2-A) = 4A ^ 2-2a +... Expansion
（1）f(1)=a+b+c=1,
f(-1)=a-b+c=0.
By subtracting, we get 2B = 1, B = 1 / 2
∴a+c=1/2.(1)
For any real number x, f (x) > = x,
ax^2-x/2+c>=0,
a> 0, and 1 / 4-4ac0, C > 0
(2) From (1) and (2), 1 / 4-4a (1 / 2-A) = 4A ^ 2-2a + 1 / 4 = (2a-1 / 2) ^ 20, so c > 0.
（2）g(x)=f（x）-mx =ax^2+(1/2-m)x+c
If G (x) is monotone on [0,1], then there is (m-1... Expansion
(1) It is shown that for any parabola (AC) ≥ 0, it has an opening in the direction of (a) ≥ 2, and it has an opening in the direction of (a) ≥ 0
If f (1) = 1, then a + B + C = 1
If f (- 1) = 0, then A-B + C = 0
By subtracting the two formulas, we get b = 1 / 2. So 4ac ≥ 1 / 4, AC ≥ 1 / 16, because a > 0, so c > 0.
（2）g(x)=f（x）-mx =ax^2+(1/2-m)x+c
If G (x) is monotone on [0,1], then (m-1 / 2) / (2a) ≥ 1 or (m-1 / 2) / (2a) ≤ 0
The solution is m ≥ 1 / (2a) + 1 / 2 or m ≤ 1 / 2
1. If f (x) ≥ x for any real number x, we get a > O and (B-1) ^ 2-4ac = 0, so 4ac > = 0, then c > = 0
When C = 0, Yi B = 1, from F (1) = a + B + C = 1, we get a = 0
So c > 0
2、f(1)=a+b+c=1，f(-1)=a-b+c=0
The solution is b = 1 / 2, C = 1 / 2-A and (B-1) ^ 2... Expansion
1. If f (x) ≥ x for any real number x, we get a > O and (B-1) ^ 2-4ac = 0, so 4ac > = 0, then c > = 0
When C = 0, Yi B = 1, from F (1) = a + B + C = 1, we get a = 0
So c > 0
2、f(1)=a+b+c=1，f(-1)=a-b+c=0
The solution is b = 1 / 2, C = 1 / 2-A and (B-1) ^ 2-4ac = 1 / 16
G (x) = f (x) - MX = ax ^ 2 + (1 / 2-m) x + 1 / 2-A, the equation of its axis of symmetry is: x = - (1 / 2-m) / 2A
Let g (x) be monotone on [0,1]
-（1/2-m)/2a=1
The solution is m = 2a-1 / 2
From a = C = 1 / 2, 1 / 4-4a (1 / 2-A) = 1
So m = 1 ﹣ put away
(1) F (1) = a + B + C = 1, f (- 1) = A-B + C = 0, two formulas add to get a + C = 1 / 2, two formulas subtract to get b = 1 / 2
For any x, f (x) = ax ^ 2 + 1 / 2x + C ≥ x, that is, ax ^ 2-1 / 2x + C ≥ 0, let H (x) = ax ^ 2-1 / 2x + C,
The formula is h (x) = a (x ^ 2-1 / 4A) - 1 / 16A + C. for any x, if h (x) ≥ 0, (when a = 0, H (x) = - 1 / 2x + C, it is a straight line
(1) F (1) = a + B + C = 1, f (- 1) = A-B + C = 0, two formulas add to get a + C = 1 / 2, two formulas subtract to get b = 1 / 2
For any x, f (x) = ax ^ 2 + 1 / 2x + C ≥ x, that is, ax ^ 2-1 / 2x + C ≥ 0, let H (x) = ax ^ 2-1 / 2x + C,
The formula is h (x) = a (x ^ 2-1 / 4A) - 1 / 16A + C, for any x, if h (x) ≥ 0, (when a = 0, H (x) = - 1 / 2x + C, is a straight line, does not meet the condition of ≥ 0, so a ≠ 0), then the quadratic function H (x) must open up, so a > 0
The minimum value of H (x) = a (x ^ 2-1 / 4A) - 1 / 16A + C is - 1 / 16A + C ≥ 0, C ≥ 1 / 16a, and C > 0 if a > 0
(2) G (x) = f (x) - MX = ax ^ 2 + 1 / 2x + c-mx = ax ^ 2 + (1 / 2-m) x + C,
Then its axis of symmetry x = (m-1 / 2) / 2a is not in (0, 1), that is, (m-1 / 2) / 2A ≤ 0 or (m-1 / 2) / 2A ≥ 1
The solution is m ≤ 1 / 2 or m ≥ A-1 / 2, so the value of M is [A-1 / 2,1 / 2]
In addition, a > 0, C > 0, a + C = 1 / 2, so 0
Find the range y = x Λ 2-3x + 4 f (x) = √ x Λ 2-2x + 4
Y = x Λ 2-3x + 4 = (x-3 / 2) & #178; + 7 / 4; ∫ (x-3 / 2) & #178; ≥ 0; ∫ y ≥ 0 + 7 / 4 = 7 / 4; so the value range is [7 / 4, + ∞); f (x) = √ x Λ 2-2x + 4 = √ (x-1) & #178; + 3; ∫ (x-1) & #178; ≥ 0; ∫ (x-1) & #178; + 3 ≥ 3; so the value range is [√ 3, + ∞). I'm glad to answer for you, skyhunt